Question: Limiting Reagent
Question 1 (Limiting Reagent)
15.00 g aluminum sulfide & 10.00 g water react until the limiting reagent is used up.
[Atomic mass: H = 1.008, Al = 26.98, S = 32.07, O = 16.00]
Here is the balanced equation for the reaction:
Al2S3 + 6 H2O ( 2 Al (OH)3 + 3 H2S (i) Which of the two reactants is the limiting reagent?
(ii) What is the maximum mass of H2S which can be
formed from these reagents?
(iii) How much excess reagent remains after the reaction is complete?
Answers:
(i) Solution for determining the limiting reagent
Determine the moles of Al2S3 and H2O aluminum sulfide: 15.00 g ÷ 150.158 g/mol = 0.099895 mol water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol
Divide each mole amount by equation coefficient aluminum sulfide: 0.099895 mol ÷ 1 mol = 0.099895 water: 0.555093 mol ÷ 6 mol = 0.0925155 which is less than 0.099895 (for aluminum sulfide)
Therefore, water is the limiting reagent.
(ii) Solution for mass of H2S formed
Now that we know the limiting reagent is water, this problem becomes "How much H2S is produced from 10.00 g of H2O and excess aluminum sulfide?"
Determine moles of 10.00 g of H2O water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol
Use molar ratios to determine moles of H2S produced from above amount of water.
(a) the H2O/H2S ratio is 6/3, a 2/1 ratio.
(b) water is associated with the two.
This means the H2S amount is one-half the water value = 0.2775465 mol.
Convert moles of H2S to grams. 0.2775465 mol x 34.0809 g/mol = 9.459 g
(iii) Solution for excess reagent remaining
We will use the amount of water to calculate how much Al2S3 reacts, then subtract that amount from 15.00 g.
Determine moles of 10.00 g of H2O : water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol
Use molar ratios to determine moles of Al2S3 that reacts with the above amount of water.
(a) the Al2S3/H2O ratio is 1/6
(b) water is associated with the 6. This means the Al2S3 amount is one-sixth the water value =
15.00 g aluminum sulfide & 10.00 g water react until the limiting reagent is used up.
[Atomic mass: H = 1.008, Al = 26.98, S = 32.07, O = 16.00]
Here is the balanced equation for the reaction:
Al2S3 + 6 H2O ( 2 Al (OH)3 + 3 H2S (i) Which of the two reactants is the limiting reagent?
(ii) What is the maximum mass of H2S which can be
formed from these reagents?
(iii) How much excess reagent remains after the reaction is complete?
Answers:
(i) Solution for determining the limiting reagent
Determine the moles of Al2S3 and H2O aluminum sulfide: 15.00 g ÷ 150.158 g/mol = 0.099895 mol water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol
Divide each mole amount by equation coefficient aluminum sulfide: 0.099895 mol ÷ 1 mol = 0.099895 water: 0.555093 mol ÷ 6 mol = 0.0925155 which is less than 0.099895 (for aluminum sulfide)
Therefore, water is the limiting reagent.
(ii) Solution for mass of H2S formed
Now that we know the limiting reagent is water, this problem becomes "How much H2S is produced from 10.00 g of H2O and excess aluminum sulfide?"
Determine moles of 10.00 g of H2O water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol
Use molar ratios to determine moles of H2S produced from above amount of water.
(a) the H2O/H2S ratio is 6/3, a 2/1 ratio.
(b) water is associated with the two.
This means the H2S amount is one-half the water value = 0.2775465 mol.
Convert moles of H2S to grams. 0.2775465 mol x 34.0809 g/mol = 9.459 g
(iii) Solution for excess reagent remaining
We will use the amount of water to calculate how much Al2S3 reacts, then subtract that amount from 15.00 g.
Determine moles of 10.00 g of H2O : water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol
Use molar ratios to determine moles of Al2S3 that reacts with the above amount of water.
(a) the Al2S3/H2O ratio is 1/6
(b) water is associated with the 6. This means the Al2S3 amount is one-sixth the water value =